Originally Posted by

**topsquark** You forgot about the cos(2x).

And check your equation. Your given solutions don't work.

This is one of the nastier ones:

$\displaystyle sin(2x) = 2cos(2x)$

$\displaystyle 2sin(x)cos(x) = 2cos(2x)$

$\displaystyle sin(x) cos(x) = cos(2x)$

$\displaystyle sin(x) cos(x) = 2cos^2(x) - 1$

$\displaystyle cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$\displaystyle cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$\displaystyle 5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions

$\displaystyle cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$

and

$\displaystyle cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $\displaystyle x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan